📚 Solving Quadratic Equations | 解二次方程
Quadratic equations appear frequently in IGCSE Mathematics. Mastering them requires a clear understanding of standard form, factoring, the quadratic formula, and the discriminant. In this article, we will walk through every essential method step by step.
在 IGCSE 数学中，二次方程频繁出现。掌握它们需要清晰理解标准形式、因式分解、求根公式和判别式。本文将逐步讲解每一种关键方法。
1. Standard Form of a Quadratic Equation | 二次方程的标准形式
A quadratic equation is any equation that can be written in the form ax² + bx + c = 0, where a, b, and c are constants and a ≠ 0. The highest power of x is 2.
二次方程是任何可以写成 ax² + bx + c = 0 形式的方程，其中 a、b、c 是常数，且 a ≠ 0。x 的最高次数是 2。
Examples of quadratic equations:
二次方程的例子：
- x² + 5x + 6 = 0 (a = 1, b = 5, c = 6)
- 2x² − 3x = 0 (a = 2, b = −3, c = 0)
- x² − 4 = 0 (a = 1, b = 0, c = −4)
Notice that the equation must be rearranged so that all terms are on one side and equal to zero.
注意：方程必须整理为所有项在一边，且等于零的形式。
2. Solving by Factoring | 因式分解法
as a product of two binomials. For example, x² + 5x + 6 = 0 factors as (x + 2)(x + 3) = 0.
当二次表达式可以写成两个二项式的乘积时，因式分解是最快的方法。例如，x² + 5x + 6 = 0 分解为 (x + 2)(x + 3) = 0。
Since the product is zero, one of the factors must be zero:
因为乘积为零，所以其中一个因式必须为零：
x + 2 = 0 → x = −2
x + 3 = 0 → x = −3
tions are x = −2 and x = −3. Always check by substituting back into the original equation.
因此解为 x = −2 和 x = −3。始终通过代入原方程进行检验。
Special cases:
特殊情况：
- Difference of squares: x² − 9 = (x − 3)(x + 3)
- 平方差：x² − 9 = (x − 3)(x + 3)
- Perfect square: x² − 6x + 9 = (x − 3)²
- 完全平方：x² − 6x + 9 = (x − 3)²
3. The Quadratic Formula | 求根公式
actoring is not possible or is difficult, use the quadratic formula. For ax² + bx + c = 0:
当因式分解不可行或困难时，使用求根公式。对于 ax² + bx + c = 0：
x = (−b ± √b² − 4ac) / (2a)
This formula works for all quadratic equations, including those with irrational or complex roots.
这个公式适用于所有二次方程，包括具有无理数根或复数根的方程。
Example: Solve 2x² + 3x − 2 = 0 using the formula.
示例：用公式解 2x² + 3x − 2 = 0。
Here a = 2, b = 3, c = −2. Substitute:
这里 a = 2，b = 3，c = −2。代入：
x = (−3 ± √(3² − 4×2×(−2))) / (2×2) = (−3 ± √9 + 16) / 4 = (−3 ± √25) / 4
So x = (−3 + 5)/4 = 0.5 or x = (−3 − 5)/4 = −2.
因此 x = (−3 + 5)/4 = 0.5 或 x = (−3 − 5)/4 = −2。
Remember to simplify the square root completely and reduce fractions if possible.
记住要完全化简平方根，并尽可能约分。
4. Completing the Square | 配方法
especially useful for finding turning points and solving equations without using the formula.
配方法将二次式重写为 a(x + p)² + q = 0 的形式。它特别适用于求顶点坐标，以及不用公式解方程。
Steps for x² + bx + c = 0:
对于 x² + bx + c = 0 的步骤：
- Take half of b: (b/2)
- 取 b 的一半：(b/2)
- Add and subtract (b/2)²
- 加上并减去 (b/2)²
- Rewrite as (x + b/2)² + c − (b/2)²
- 重写为 (x + b/2)² + c − (b/2)²
Example: Solve x² + 6x − 7 = 0 by completing the square.
示例：用配方法解 x² + 6x − 7 = 0。
(x + 3)² − 9 − 7 = 0 → (x + 3)² − 16 = 0
(x + 3)² = 16 → x + 3 = ±4 → x = 1 or x = −7
When the coefficient of x² is not 1, first factor it out before completing the square.
当 x² 的系数不是 1 时，先提出该系数，再进行配方。
5. The Discriminant | 判别式
The expression b² − 4ac is called the discriminant. It tells us the number and type of roots without solving the equation.
表达式 b² − 4ac 称为判别式。它无需解方程就能告诉我们根的数量和类型。
| Discriminant (Δ) | 判别式 (Δ) | Roots | 根的情况 |
|---|---|---|---|
| b² − 4ac > 0 | 大于 0 | Two distinct real roots | 两个不同的实数根 |
| b² − 4ac = 0 | 等于 0 | One real repeated root | 一个实数重根 |
| b² − 4ac < 0 | 小于 0 | No real roots | 无实数根 |
Example: For x² + 2x + 5 = 0, discriminant = 4 − 20 = −16, so there are no real roots.
例如：对于 x² + 2x + 5 = 0，判别式 = 4 − 20 = −16，因此没有实数根。
This is often tested in non-calculator papers, so practice mental calculations.
这在非计算器试卷中经常考查，所以要多练习心算。
6. Graphs and Roots | 图像与根
equation ax² + bx + c = 0 correspond to the x-intercepts of the parabola y = ax² + bx + c.
二次方程 ax² + bx + c = 0 的解对应抛物线 y = ax² + bx + c 与 x 轴的交点。
If the discriminant is positive, the graph crosses the x-axis twice. If it is zero, the graph touches the x-axis at one point. If it is negative, the graph does not intersect the x-axis.
如果判别式为正，图像与 x 轴相交两次。如果为零，图像与 x 轴相切于一点。如果为负，图像不与 x 轴相交。
of symmetry of the parabola is x = −b/(2a), which is also the x-coordinate of the vertex.
抛物线的对称轴是 x = −b/(2a)，这也是顶点的 x 坐标。
e intercepts are (1,0) and (3,0). The vertex lies on x = 2, and substituting gives y = −1.
示例：画出 y = x² − 4x + 3 的草图。根为 x = 1 和 x = 3，所以交点为 (1,0) 和 (3,0)。顶点位于 x = 2，代入得 y = −1。
7. Common Mistakes and How to Avoid Them | 常见错误与避免方法
Students often make errors when solving quadratic equations. Here are the most common ones and their fixes.
学生在解二次方程时常犯错误。以下是最常见的问题及解决方法。
- Forgetting to write the equation in standard form. Always rearrange to ax² + bx + c = 0 before factoring or using the formula.
- 忘记将方程写成标准形式。在因式分解或使用公式前，务必整理为 ax² + bx + c = 0。
- Losing the negative sign. When substituting into the formula, include negative signs carefully.
- 丢失负号。代入公式时，小心处理负号。
- Dividing by a variable. Never divide by x unless you know x ≠ 0; you may lose a root.
- 除以一个变量。除非你知道 x ≠ 0，否则绝不要除以 x；否则可能丢失一个根。
- Simplifying √ incorrectly. For example, √9+16 = √25 = 5, not √9 + √16 = 7.
- √ 化简错误。例如，√9+16 = √25 = 5，而不是 √9 + √16 = 7。
Always check your answers by substituting them back into the original equation.
始终通过代入原方程来检验答案。
8. Practice Problems | 练习题
Try these questions to test your understanding. Solve each equation and check your answers.
尝试以下问题来测试你的理解。解每个方程并检查答案。
- x² − 7x + 12 = 0
- 2x² + 5x − 3 = 0
- x² − 16 = 0
- x² + 8x + 15 = 0
- 3x² − 6x + 2 = 0 (give answers to 2 decimal places)
- x² + 4x + 4 = 0
Answers: 1) x = 3, 4 2) x = 0.5, −3 3) x = ±4 4) x = −3, −5 5) x ≈ 1.58, 0.42 6) x = −2 (repeated)
答案：1) x = 3, 4 2) x = 0.5, −3 3) x = ±4 4) x = −3, −5 5) x ≈ 1.58, 0.42 6) x = −2（重根）
Make sure you can also solve by completing the square and by using the quadratic formula for each of these.
确保你也能用配方法和求根公式解这些题。
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